If $z_{1}, z_{2}, z_{3}, z_{4}$ are represented by the vertices of a rhombus taken in the clockwise order then

z_{1} - z_{2} + z_{3} - z_{4} = 0

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z_{1} + z_{2} = z_{3} + z_{4}

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a m p \frac{z_{2} - z_{4}}{z_{1} - z 3} = \frac{π}{2}

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a m p \frac{z_{1} - z_{2}}{z_{3} - z_{4}} = \frac{π}{2}

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Solution

The correct options are
A $z_{1} - z_{2} + z_{3} - z_{4} = 0$
C $a m p \frac{z_{2} - z_{4}}{z_{1} - z 3} = \frac{π}{2}$
Rhombus, being a parallelogram, the diagonals bisect each other.
So, $\frac{z_{1} + z_{3}}{2} = \frac{z_{2} + z_{4}}{2}$
$\Rightarrow z_{1} + z_{3} = z_{2} + z_{4}$
Also, another property in a rhombus says that the diagonals are perpendicular to each other.
$\Rightarrow a m p \frac{z_{2} - z_{4}}{z_{1} - z_{3}} = \frac{π}{2}$

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